hdu 3823 素数打表+vector可变长数组 好烦的一道好题。。。。。

http://acm.hdu.edu.cn/showproblem.php?pid=3823

Prime Friend

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2855    Accepted Submission(s): 580

Problem Description

Besides the ordinary Boy Friend and Girl Friend, here we define a more academic kind of friend: Prime Friend. We call a nonnegative integer A is the integer B’s Prime Friend when the sum of A and B is a prime.
So an integer has many prime friends, for example, 1 has infinite prime friends: 1, 2, 4, 6, 10 and so on. This problem is very simple, given two integers A and B, find the minimum common prime friend which will make them not only become primes but also prime neighbor. We say C and D is prime neighbor only when both of them are primes and integer(s) between them is/are not.

 

Input

The first line contains a single integer T, indicating the number of test cases.
Each test case only contains two integers A and B.

Technical Specification

1. 1 <= T <= 1000
2. 1 <= A, B <= 150

 

Output

For each test case, output the case number first, then the minimum common prime friend of A and B, if not such number exists, output -1.

 

Sample Input

22 43 6

 

Sample Output

Case 1: 1Case 2: -1

题解

题意：  即求一个数，能使a，b与之相加后，成为素数，并且a与b之间没有其他的素数。

vector开变长数组

         初始化：vertor<int> a[100];               //即开的是二维数组，可以包含100行，每行的长度可以不同。

 

                     a[2].push_back(i);                    //将i放入第二行的尾部

 

                     a[2].size();                           //获得第二行的长度

心路历程：...

一开始写了一遍。。运行不了。。debug发现prime数组开小了

开大。。。提交。。。超内存...

max改成  ? :  。。降了一半

cin .cout改成scanf..  没变化。。

把isprime()拉到主函数里来。。。 没变化。

发现素数打表有一个地方写错。。。改正了。。没变化。。

把素数的数组改成bool（一开始用Int）。。。好了一点。。。

最后不用G++提交了。。。换了C++提交。。。然后就过了。。。。。。。好气哦。。。做了这么久的题

//hdu 3823#include <iostream>#include <cstring>#include <cstdio>#include <string>#include <algorithm>#include <vector>using namespace std;typedef long long ll;//题意：给出两个数字a，b。使a + x, b + x都是素数，并且它们之间没有素数。求出这样的最小的x。//http://www.cnblogs.com/yumao1006/archive/2012/08/30/2664591.html/*暴力打表,找出20000000以内的素数，求出相邻素数的差，存放在vector的数组中*///学习：：把max函数换成?:能省一半内存。。。。//学习：：把cin cout 换成scanf，能省一半时间。。。//一开始，素数打表那里条件写成i < maxn 。。。错了好多次啊。。const int maxn = 20000000 + 5;//在这里面素数的差才可能在150以内bool pp[maxn];//筛选法找素数int prime[maxn/2];//存放素数vector <int>vec[155];void isprime(){memset(pp, 0, sizeof(pp));pp[1] = 1;for (int i = 2; i*i < maxn; i++){if (!pp[i]) //0是素数{for (int j = i + i; j < maxn; j+=i){pp[j] = 1;}}}}int main(){isprime();//找素数个数，并把素数都放在数组prime[15000]中int times = 0;for (int i = 2; i < maxn; i++){if (!pp[i]){prime[times++] = i;}}for (int i = 0; i < times-1; i++){int tmp = prime[i + 1] - prime[i];if (tmp <= 150)vec[tmp].push_back(prime[i + 1]);}int t, cnt=0;int a, b;scanf("%d", &t);while (t--){//cin >> a >> b;scanf("%d%d", &a, &b);int abss = abs(a - b);//两个数的差一直都不变int maxx = a > b ? a : b;//一开始的最大int res = 0;//加上x后的最大for (int i=0;i<vec[abss].size();i++){if (vec[abss][i] >= maxx){res = vec[abss][i]; //找到一个能满足的最小的就行了break;}}printf("Case %d: ", ++cnt);//printf(res ? "%d\n" : "-1\n", res - maxx);if (res == 0)puts("-1");else//cout << res - maxx << endl;printf("%d\n", res - maxx);}system("pause");return 0;}